# Question #ee6eb

Apr 21, 2015

The balloon will get bigger if you increase temperature while keeping pressure and number of moles constant.

According to Charles' Law, when pressure and number of moles are kept constant, the volume an ideal gas occupies is directly proportional to its temperature.

The higher the temperature, the bigger the volume. Likewise, the lower the temperature, the smaller the volume.

Mathematically, this is written as

${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}$, where

${V}_{1}$, ${T}_{1}$ - the volume and temperature at an initial state;
${V}_{2}$, ${T}_{2}$ - the volume and temperature at a final state.

According to the above equation, the volume of the balloon will be

${V}_{2} = {T}_{2} / {T}_{1} \cdot {V}_{1}$

Since ${T}_{2} > {T}_{1}$, the ratio between the two temperatures will be greater than 1

${T}_{2} / {T}_{1} > 1 \implies \textcolor{g r e e n}{{V}_{2} > {V}_{1}}$